Non-commutative Valuation Rings and Their Global Theories （非可換付値環とその大域理論）

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(1)Non-commutative Valuation Rings and Their Global 'Theories. 2007. The Joint Graduate School in the Science of School Education. Hyego University of Teacher Education. Department of Science School Education (Algebra). Santi Irawati.

(2) ACKNOWLEDGEMENT I would like to express my deep gratitude to my advisor, Prof. Dr. Hidetoshi. Marubayashi for his guidance during my study. He gave me invaluable ideas which made this dissertation possible.. I would like to thank Prof. Dr. Sri Wahyuni, as my home advisor, for her guidance and helpful support in getting the opportunity for studying abroad.. I also would like to thank JSPS RONPAKU program for the financial support which gave me a chance to study ring theory under good mathematicians in Japan.. I am gratefu1 to the Professors and staffs of Naruto University of Education especially ofthe department ofMathematics, for their care and hospitality..

(3) CONTENTS Introduction. 1. CHAPTER 1 Some elementary properties. 1.1. Someelementarypropeniesoforders 1.2. SomeelementarypropertiesofDubrovin valuation rings 1.3. Some elementary properties of Pnifer orders. -------t-------------. --------i------------. ----i----------------. CHAPTER 2 On R-ideals of a Dubrovin valuation ring R 2.1. Structureofv-ideals 2.2. R-ideals of a Dubrovin valuation ringR. ---------------------. 5 5. 7 11. 13 13 16. CHAPTER 3 A characterization of fully bounded Dubrovin valuation rings. .......H..."..."" 22. 3 . 1 . Fully bounded Dubrovin valuation rings. "".........-.."" 23. 3.2. Examples. ......."."....."" 28. CHAPTER 4 Non-eommutative v-Bezout rings 4.1. Non-commutativev-Bezoutrings. 30 30. 4.2. Examples. 33. C[HAPTER 5 Overrings of non-commutative PrUfer rings 5. 1 . 0verrings of PI Prtifer rings. •••...."........"" 36 ................."" 36. 5.2. Pri me ideals of ovenings of a PI PrUfer ring. ..."..-......."" 39. References. ............".."." 41. satisfying a polynomial identity.

(4) Introduction During the last twenty years the theory of non-comnutative valuation rings has been developed by many authors for different reasons. The main progress in the general. theory has been made after N. I. Dubrovin introduced his new type of valuation rings which are called Dubrovin valuation rings. These rings are not only defmed for division rings but also for sirnple Artmian rings especially for central sirnple algebras. As we know, there are three types of non-commutative valuation rings, which are called total valuation rings, invariant valuation rings and Dubrovin valuation rings.. Let K be a division ring. A subring Vof K is called total valuation ring of K if. for any non-zero element a E K, either a E Vor a-i E V. A total valuation ring Vof a division ring K is called an invariant valuation ring if aV = Va for all a E K. An order R. in a simple Artinian ring e is called a Dubrovin valuation ring of 2 if R is Bezout and. R!J(R) is simple Artmian, where J(R) is the Jacobson radical of R. We see that every invariant valuation ring and every total valuation ring V is clearly a Dubrovin valuation ring. However, the converse is not necessarily true.. In this thesis, we study about non-commutative valuation rings in particular about Dubrovin valuation rings and their global theories, say PrUfer imgs. Moreover, we give some examples of v-Bezout rings which are the generalization of commutative. GCD-domains. / In Chapter 1, we give some elementary properties of non-commutative valuation imgs, which are used in the next Chapters. We refer to [MMU] for details conceming with orders, Dubrovin valuation rings, PrUfer orders and primary ideals.. Let R be an order in a ring 2. A right R-submodule I of 2 is called a right Rideal of e if (i) I A U(2) # SZ5, where U(2) is the unit group of 2 and (li) there exists c. E U(e) such that cl s R. A left R-ideal of 2 is defmed similarly. A right and left Rideal is called an R-ideal. For a right R-ideal I of Q, we set 0,(I)= {gE 2l Ig {;;I},. 1.

(5) the right order ofI and O, (I)= { q E 2] gl gl }, the left order of L An element c in e is called a right stabilizing element of R if cR is an R-ideal and we denote by r-st(R) = { c E e 1 cR is right stabilizing}. We say that c is stabilizing if cR = Rc and denote by. st(R) = {cE2 1c is stabilizing}. For any ideal I of a ring R, we denote by V-]Z-=. A{Pl P E Spec(R) withP2I} theprime radical oflwhich is a semi-prime ideal. An idealA ofR is called ri ht .vf7-primary if aRb gA, where a, bER, implies either aE A or bE V-i7 . Similarly, left ptmary ideals are deiined. In [BMU], they have described all right primary ideals of R.. In Chapter 2, we investigate the structure of all R-ideals by usage of stabilizing. elements and prirnary ideals by using some results from [BMU]. IfI is an R-ideal and I is not fmitely generated as a right R-ideal such that 0, (I) = S = O, (I) and suppose that. .I(S) is Archimedean, it is proved that I :cA for some cE st(S) and A, a right and left. J(S)-primary ideal (see Theorem 2.2.3). In the case 2 is fmite dimensional over its center, we obtain: (1) If I is finitely generated as a right R-ideal, then I = cR = Rc for. some cG st(R), (2) IfI is not finitely generated as a right R-ideal such that J(S) is. Archimedean, thenI= cA = Ac for some cE st(S) and A, a right and left J(S)-pmbary ideal, (3) Ifl is not fmitely generated as a right R-ideal such that .1(S) is limit prime, then. I is one of the following three; I= cS = Sc for some cE st(,S), l = cJ(R) = 1(R)c for. some cE st(R) andI= nc2Rz for some cz E st(R2), where RA= Rp, and P2 runs over. '. all Archimedean prime ideals with P2c J(S) (see Proposition 2.2.4). Furthermore, a counter example is given to show that Proposition 2.2.2 (2)(a) is not necessarily held if e is infmi/ te dimensional over its center.. A ring is called right aeft? bounded if any essential right (left) ideal contains a. non-zero (two-sided) ideal. A img isjust called bounded if it is both right bounded and. left bounded. Let Sbe aring. We say that Sisfully bounded if S/P is bounded for any prime ideal P of S. Let R be a Dubrovin valuation ring in a simple Artinian ring 2 and let P E G-Spec(R), the set of all Goldie prime ideals ofR, with P # J(R) and set P, =. 2.

(6) A{ Pi1 Pa E G-Spec(R) with P, D P}. Then, in [BMO,(6)], they have shown that the. following four cases only occur: ' (1) P is lower 1imit, i.e.,P= P, . Otherwise, P, )P is aptme segment.. (2) P, D P is Archimedean. (3) P, DP is simple. (4) P, D P is exceptional, i.e., there exists a non-Goldie prime ideal C such that P, D. C)P.. '. '. In Chapter 3, we investigate those results under an additional condition that R is fully bounded. It is shown that for a Dubrovin valuation ring R of a simple Artmian img 2, R is fully bounded iff (1) and (2) only hold.. Moreover, for any regular element c in J(R), we defme P(c) = A{ P2[ Pz e GSpec(R) with c E P, }, a Goldie prime ideal. R is called locally invariant if cP(c) = P(c)e for any regular element c in J(R). Let R be a Dubrovin valuation ring of a simple Artinian img 2. It is shown that R is fully bounded if and only if it is Iocally invariant.. If 2 is of-fmite dimensional over its center, then R is always fully bounded. In the end of this chapter, we give several examples of fully bounded Dubrovin vaiuation rings of 2 with inf/mite dimension over its center.. In Chapter 4, we study non-commutative GCD-domains. An Ore domain S is called right aeft? v-Bezout if I, is a principal for any fmitely generated right ideal I of S.. S is said to be v-Bezout if it is right v--Bezout as well as left v-Bezout. This img is a non-. commutative version of a commutative GCD-domain. In the commutative imgs, [Gi] proved if R is a GCD domain, so is R[x]. Inspired by [Gi], we prove if Vis a total valuation ring of a division ring K, then R = l7[x",6 1r E Q,] is v-Bezout where Q, the set of non-negative rational numbers, 6: Q,---> Aut(l7) is defmed by 6 (r + s) =. 3.

(7) 6(r).6(s) for any r, s E Qo,. and the multiplication in R is. defined by x'a =. xr for any aE JZ andrE Qo•. In Chapter 5, we study prime ideals of any overring of a non-commutative PI. Pnifer ring. We defme I-i= {gE21 Igl gl}), the inverse ofL Following [AD],R is called right Profer if for every finitely generated right R-ideal L I"iI = R, II-' = O, (I)•Left Profer rings are defmed similarly. In [D2], he proved that any prime ideal of a PI Prdfer ring is localizable. In the case when prime rings satisfytng a PI, [Mo] studied. PI PrUfer rings under some conditions. By using some results in [D2] and [Mo], we shall prove if S is an overring of a prime Goldie ring R and suppose that R is PrUfer satisfying. a polynomial identity, then Spec(S) = {PS 1 P E Spec(R) with PS c S}and S = AR. , where P runs over all PE Spec(R) with PS a S.. 4. 6(r)(a).

(8) CHAPTER 1 Some elementary properties. In this chapter, we give some elementary properties of orders, non-commutative. Dubrovin valuation rings and PrUfer orders. We refer to [MMU] and [MR] for details concerning with orders.. 1.1. Some elementary properties oforders. '. In this section, we give some defmitigns, notations and elementary properties of orders. For a ring R, we denote by U(R) the set of all units ofR and by C. (O) the set of all regular elements (that is, non-zero divisors) of R.. Let C be a multiplicatively closed subset of a img R. We say that R satiisfies the right Ore condition with respect to C or that C is called a right 0re set of R if, for any. aE R and cE C, there exist bE R and dE C such that ad = cb. If C g C. (O), then it is called a regular right Ore set of R. Similarly, we can define a (regular) left Ore set of R. If C is a (regular) right and Ieft Ore set of R, then it is simply called (regular) Ore set. ofR• Let C be a regular right Ore set of a ring R. An overring T of R is called the right euotient ring of R with respect to C if. (i) cE U(T) for any cE C and (li) for any g E T, there exist aE R and cE C such that g=ac-i .. We denote the ring T by R. . We note that ,for a multiplicative subset C of R with C g. C.(O), the right quotient ring R. ofR with respect to C exists if and only if C is a regular right Ore set ofR ([MR, Chap. 2]). A subring R of a ring 2 is called a right order in 2 if 2 is the right quotient ring. ofR with respect to C. (O), and sometimes we denote the ring 2 by e(R). In particular,. R is aright order in 2 if and only if C.(O) is aright Ore set of R. Similarly, we can defuie a left order in 2 and a ring which is both a right and left order in 2 is called a.n. order in 2•. 5.

(9) Let R be a ring and let Mbe a right R-module. An R-submodule L ofMis called. essential if L A N # O for any non-zero R-submodule N of M. By Zorn's lemma, we note that for any R-submodule L of M, there exists an R-submodule L ' of M such that L fi L' = O and L eD L' is essential in M([MR, (2.2.2(v))]). If a right ideal I ofR is an es sential R-submodule ofR, then I is called an essential right ideaL A right R-module U. is said to be uoform if, for any non-zero R-submodule U, and U2 of U, Ui A U2 4 O, that is, any non-zero R-submodule of Uis an essential R-submodule of U.. A right R-module M is said to havefinite Goldie dimension if it contains no infinite direct sum of non-zero R-submodules. For any subset X of R, we set rR (X) = { a E R l Xa = O} and call it the right annihilator of X. Sometirnes we denote r. (X) by roo. The left annihilator of Xis defmed similarly. A ring R is called a right Goldie ring. if R satisfies the ascending chain condition (acc) on right annihilators and has fmite Goldie dimension as a right R-module. A left Goldie ring is defmed similarly and R is called a Goldie ring ifR is a right and Ieft Goldie ring. We have the following property. ([MMU, (1.1)]).. Theorem 1.1.1. Let R be a ring. Then thefollowing is equivalent: ld R is a (semi)-prime right Goldie ring. (iO R has a right quotient ring 2 which is (semi)-simple Artinian, that is, R is a right. order in a (semO-simp le Artinian 2.. Let R be an order in a ring 2. A right R-submodule I of 2 is called a right Rideal of 2 ifIA U(2) #/ and there exists cE U(e) such that cl g R. A right R-ideal I of e is said to be integral ifI g R. Similarly, we can defme a left R-ideal of 2. A right and left R-ideal is called an R-ideal.. Let R an order in a ring 2. For any subsetsX and ]Y of 9, we set (X :Y),=. {qEel gYgX, (X :Y), ={gE21 Yg gX } and X"'={gE2l XgX gX }. For a ' we called it the right right R-idealIof 2, we set 0,(I)= (I :I).' { qE 2I Ig gl } and ' A right and left R-ideal is called an Rorder of L The left order of i is defmed sirnilarly. ' ideal. Then we have the following lemma ([MMU, 1.2]).. 6.

(10) Lemma 1.1.2. Ir R is an order in a ring 2 andl is a right R-ideal of 2 then (1) O,(I)andO, (I) are orders in 2,. (2) I is a left O, (I)-ideal andarightO,(I)-ideal, and (3) (R :I), is a leftR-ideal andaright O, (I)-ideal.. 1.2. Some elementary propenies ofDubrovin valuation rings in this section, we give some elementary characterizations of Dubrovin valuation imgs and its ideal theory.. Let D be a division ring, (G, +) be a totally ordered group and let U(D) be the set of all units in D. A suijective mapping v: U(D) -> G is called a valuation on D if it is satisfyjng ([Sc]):. (1) For any a, b E D, v(ab) = v(a) + v(b)• (2) v(a + b) ) min { v(a), v(b) } if b # -a.. Ifv is a valuation on D, then V= {aEU(D) 1 v(a) 2O' } u {O}. Then Vis an invariant valuation ring. In this case G is called value group of V.. A ring R is called right Bezout if any fmitely generated right ideal of R is principal. The left Bezout is defined similarly. A ring R is called Bezout if it is right and left Bezout.. Let R be a subring of a ring S. R is called a right n-chain ring in S if for any n+1. elements ao, ai,..., a. in S, there is anisuch that a,E2k.,akR.A right n-chain ring in itseif is called a right n--chain ring. A left n-•chain ring is defined similarly. An n-. chain ring is a right and left n-chain ring. Then we have the following propenies ([MMU, (5.8), (5.11) and (5.12)]).. Lemma 1.2.1. LetR be a semi-simple ring, that is, J(ll) = O. ThenR isArtinianf and only f R is a right n-chain ringfor some n.. 7.

(11) Theorem 1.2.2. Let R be a subring of a simple Artinian ring 2. Then thefollowing conditions are eguivalent.'. (1) R is a Dubroyin valuation ring of 9• (2) R is a local semi-hereditaiv order in 2• '. (3) R is a local Bezout order in 2. (4) R is a local n-chain ring in efor some n with d(R? -> n, where R = R/J(R).. Corollary 12.3. Let R be a Dubrovin valuation ring of 2 and let P be aprime ideal of R. ,lr R/P is aprime Goldie ring, then R/P is qlso a Dubrovin valuation ring of its classical guotient ring.. Lemma 1.2.4. ([MMU, (6.3)]). LetR be aDblbrovin valuation ring of 2 and let T2g Ti be right R-submodules of 2 such that (1) T, is regular and (2) there exists a subring. S of O,(T,)= {gE2lgT2 gT2} such thatfor aay regular elements t, and t,ET, there is a regular element t e T, with St, +St2 g St• Then eitherTi= T2 or tiJ(R) ? T2 for some regular element ti E Ti•. By using Lemma 1.2.4, we have the following Proposition (IMMU, (6.4)]). Proposition 1.2.5. Let R be a Dubrovin valuation ring of 2 and let S be a Bezout order in 2. Then the set of regular left S- and right R-submodules of e is linearly ordered LIy inclusion. Inparticular, the set of all R-ideals of 2 is linearly ordered by inclusion.. Let P be a prime ideal of a ring S If C(P) = {cE S1c: regular mod P} is a. '. regular Ore set ofS then the quotient ring S.(.) of S with respect to C(P) is denoted by. S. and is called the localization of S at P. Let R be a Dubrovin valuation ring of a simple Artinian img e and let S be an overring of R. Then J(S) g J(R) and S is local. ([MMU, (5.3)]). Combining Lemma 1.2.1 and Theorem 1.2.2, we have ([MMU, (6.6)]). 8.

(12) Theorem 1.2.6. Let R be a Dubrovin valuation ring of e and let S be an overring of R. (1) R-'. = RfJ(S) is a Dubrovin valuation ring of S = S/J(S)• ,. (2) S is a Dubrovin valuation ring of 2 and J(S) is aprime ideal of R. (3) C. (J(S)) is a regular Ore set of R andS= RJ(s) •. The converse of Theorem 1.2.6 (1) also holds ([MMU, (6.16)]) as following.. '. -Proposition 1.2.7. Let S be a Dubrovin tvaluation ring of 2 and let R be a Dubrovin valuation ring of S= S!J(S)• Then the setR = {rES I [r+.I(S)] E R } is a Dubrovin valuation ring of 2•. Let R be a Dubrovin valuation ring of 2. Then O, ( J(R)) = O, ( J(R)) = R by [MMU, (6.8)], which implies the following Lemmas ([MMU, (6.9) and (6.10)]).. Lemma 1.2.8. Let R be a Dubrovin valuation ring of e , A be an R-ideal of 2 and S == O,. (t4). Then thefollo;ving are eguivalent:. (1) A isprincipal as a right S-ideal•. (2) A-iA =s (3)A )AJ(S).. Lemma 1.2.9. LetR be a Dubrovin valuation ring of 2 and letA be an R-ideal of e. Then 0. (t4 ) = O, (A -' ) and O, (A) = O, (A -') .. ' Combining Lemma 1.2.4 and Lemma 1.2.9, we have. Propositien 1.2.le ([MMU, (6.13)]). Let R be a Dubrovin valuation ring of 2 and let. A be an R-ideal of 2. SetS =O, (A) and T= O, (A)•. (1) Av= (S: (S :A)i), =A"= 'A =(T: (T:A),)i andA*=A-i -'.. 9.

(13) (2) A"" =A' and (A -i )" = A-i . (3) JrA is not aprincipalrightS-ideal, then A-iA = J(S) andJ(S) is not aprincipal right S- ideal.. (4) IfA cA* , thenA* " cS andA = cJ(S)for some regular element c EA*• In particular, A =A* J(S)•. Let Ibe a right R-ideal and Iet S= O, (I). We define I" = A cS, where c runs over all elements in e with cS R J Sirnilarly, for any left R-ideal L with T = O, (L), we. defme *L = A Tc, where c runs over all elements in 2 with Tc p L. The following proposition is established by a standard method ([MMU, (6.I 1)]).. Proposition 1.2.11. Let R be a Dubrovin valuation ring of 2 and let I be a right R-. ' (1) IgI*. (2) (I*)* = I*.. (3) (cD" = cl* for aigy c E U(e)•. (4) (cD -' =I'ic-ifor aay c E U(2)•. Let R be a Dubrovin valuation ring of 2. Then by [MMU, (6.8)], O. (1(R)) = O, (J(R)) = R. If J(R) is not principal as O. (J(R))-ideal, then J(R)* = R = J(R)-i by. [MMU, (6.12)]. A prime ideal P of a ring R is called Goldie if RIP is a Goldie ring. By [MMU, (6.8)], if R is a Dubrovin valuation ring of a simple Artmiaii ring 2 and S is an overring. ofR then J(S) is a Goldie prime ideal of R, which is localizable and R. is a Dubrovin valuation ring with J( R. ) A R = P ([MMU, (14.5)]). We denote by $(R) the set of all ovenings of a img R, Spec(R) the set of all prime ideals of R, and G-Spec(R) the set of. all Goldie prime ideals of R. By [MMU, (6.7) and (14.5)], we have the following. correspondence. 10.

(14) Proposition 1.2.12. Let R be a Dubrovin valuation ring of a simple Artinian ring 2. Then there exists a one-to-one correspondence between $(R) and G-Spec(R).. Let R be a Dubrovin valuation ring of a sirnple Artinian ring 2. Then the intersection of Goldie prime ideals ofR is also a Goldie prime ideal ([BMO, (1)]). Proposition 1.2.13. Let R be a Dubrovin valuation ring of a simple Artinian ring 2. andlet P, E G-Spec(R). ThenP=A P, is also in' G-Spec(R).. Let R be be a Dubrovin valuation img of a division ring K. The following Lemma ([MMU, (8.13)]) gives a criterion ofR to be a total valuation ring.. Lemma 1.2.14. Let R be be a Dubrovin valuation ring of a division ring K. Then R is total jf and only f ii = Rll(R) is a division ring.. 1.3. Some elementary properties ofPrUfer orders In this section, we give some properties of PrUfer orders. Let 2 be a semi-simple. Artinian ring and let R be an order in 2, that is, R is a semi-prime Goldie ring. R is called a right aeft? Profer order in Q if any fmitely generated right (left) R-ideal is a. progenerator of Mod-R (R-Mod), that is, projective and a generator of Mod-R (RMod). A Pr(]"er order is a right and left Pnifer order. By [MMU, (2.5)], a right PrUfer. order in asemi-simple Mmian img is left Prdfer order. • Let R be a Priifer order in a semi-simple Artinian img 2 and let S be an overring. of R, that is, R g S g e. It is clear that S is an order in 2. It is proved in ([MMU, (2.6)]) that S is also PrUfer.. Proposition 1.3.1. An overring of a Pr(ifer order in a semi-simp le Artinian ring is also a Profer order.. Let A be an ideal of a Prttfer order R in a simple Artmian ring 2. Then any element of C(A) = {cER 1 c is regular modulo A} is regular ([MMU, (22.6)]). In the. 11.

(15) caseA is maximal such that RIA is a semi-simple Artinian ring, then C(,4) is an Ore set. ofR and R. is a Dubrovin valuation ring of2 ([MMU, (22.7)]).. Proposition 1.3.2. LetA be an ideal of a Pofer order in, a simple Artinian ring 2. Then C(A) consists of regular elements of R.. Theorem 1.3.3. Let R be a Profer order in a simple Artinian ring 2 and letA be an ideal of R such that RIA is a semi-simp le Artinian ring. (1) C(`4) is a regular Ore set of R.. (2) gr A is a maximal ideal of R, then R. is aDubrovin valuation ring of2.. Dubrovin has proved the following property of Prdfer order ([D2, (4)]). Proposition 1.3.4. Let R be a Profer order in a simple Artinian ring e and let S be a. Dubrovin valuation ring of e containing R. Then P = J(S) A R is aprime ideal of R sueh that C(P) is a regular Ore set of R andS= R.•. The following Proposition is proved by [Mo, (3.1)]. Proposition 1.3.5. Suppose S is a Dubrovin valuation ring and EII is an order inS =. SILI(S). ThenR" {rES I r+J(S) E 611} is Proferfand only f Sl is Profer.. 12.

(16) CHAPTER 2 On R-ideals of a Dubrovin valuation ring R. Throughout this Chapter, we denote by R a Dubrovin valuation ring in a simple Artinian ring 2. We use "c" or `5" for proper inclusion and "g" or "p" for inclusion.. For any subset X and Yof e, we set (X:Y),={gE2IgYgX} and (X:Y),= {geel Yq gX }. For an R-ideal I, we set I. = (R :(R :I),), and ,I = (R :(R:I).)i• Iis called av-ideal if I. =I= ,I• In Section 2.1, we give some structures of v-ideals related to the properties of Jacobson radical.. In Section 2.2, it is described the structures of all R- ideals by usage of stabdizing elements and primary ideals.. We refer to [MMU] and [BMU] for details conceming with Dubrovin valuation imgs and primary ideals.. 2.1. Strueture ofv-ideals. For an R-ideal L we set 0,(I)= {qE21Iq gl }, the right order of I and Oi (I)= { g E 2 l gl gl }, the left order of L Then we have the following. Lemma 2.1.1. Let S be aproper overring of R. Then (1) (R : S), = J(,S[). (2) (R:1(S)), =S. proofr (1) It is clear that (R :S), pJ(S). If (R :S), DJ(S), then (R :S), = (R :S),S= S because (R : S)t is an ideal ofR and S= RJ(s), a contradiction.. (2) It is clear that (R:J(S)), R S. To show the converse inclusion, fifst assume that J(S) = sS = Ss for some s E J(,S). Then we have (R : J(S)), = (R : S), s-i = J(ED sMi = S by (1). Next, assume that J(S) is not fmitely generated as a one-sided S-ideal. Then we. 13.

(17) have J(S) = J(S)2 and O, (J(S)) = S by [MMU, (6.8)] and Lemma 1.2.8, and it follows that (R :J(S)),g(S :J(S)),= O, (J(S)) = S. Hence (R :J(S)), = S.. A prime ideal P of R is said to be Goldieprime if R/P is a Goldie img. By Theorem 1.2.6 and Proposition 1.2.12, P is Goldie prime if and only ifR. exists and is a Dubrovin valuation ring. We note that J(S) is always Goldie prime for any overring S of. R.. Lemma 2.1.2. Let I be an R-ideal and set S = O. (I) and T= O, (I) • Then (1) ff I isfinitely generated as a right R-ideal, then (R :J),I= R• (2) l7" I is notfinitely generated as a right R-ideal, then (R :I),I = J(S . In. Particular, (R :I),I is Goldieprime.. Progf: (1) It is clear.. (2). IfI= aS for some a E I, then S# R by assumption, and so (R :I)," (R :aS),= (R :S), a-i = J(S) a-i by Lemma 2.1.1(1). Hence we have (R :I),I= J(S) a-i aS = J(S). IfI is not'. fmitely generated as a'. right S-module, then I = I J(S) by Lemma 1.2.8. It. follows by Lemma 2.1.1(2) that (R:I), = (R:IJ(S)), = ((R:J(S))i:I)i = (S:I), =I-i (:" {xEel Ixl gl }). Thus By Proposition 1.2.10 (3), J(S) = I-iI = (R :I),L. An element c in e is called a right stabilizing element of R if cR is an R-ideal and.we denote by r-st(R) = { c E e i cR is right stabilizing}. We say that c is stabilizing is cR = Rc and denote by st(R) = { c E 2 ] c is stabilizing}.. If S is a Noetherian Prdfer order in a simple Artmian ring, i.e., a Dedekind ring, theR any ideal is always a v-ideal, because it is projective. However, in non-Noetherian case, this is not necessarily to be held. In the case of Dubrovin valuations rings, this. depends on the properties of Jacobson radical, as it will be seen in the fo"owing proposition which is used in section 2.2.. 14.

(18) Proposition 2.1.3. , '. (1) lr J(R) = xR = R)cfor somex e R, then a7!y R-ideal is a v-ideal. (2) 3r J(R) -= J(R)2, then tcJ(R)/ce r-st(R)? is the set of all non v-ideals.. ProqC Let I be an R-ideal with I c I. . Then I c- cul(R) c- I, for some regular element. aE I. by Lemnia 1.2.4. So I. g (a.I(R)).= (axR).= axR g I..= I. • Thus I, = axR g aR g I, , and we have J(R) = xR = R, a contradiction. Hence I = I, , and similarly we. havel= I. v. (2) By [MMU, (6.8), (6.12)] and Proposition 1.2.11, we have (cJ(R)),= c(J(R)).= cR D cJ(R) for any cE r-st(R), and so cJ(R) is not a v-ideal. Conversely,'. let I be an R-ideal. with I c I..Then, by Lemma 1.2.4, I g cJ(R) g I, for some regular element eE I,. So J,= (eJ(R)),= cR. Thus cE r-st(R). Now we shall show that I= cJ(R). To prove this assume, on the contrary, that I c cJ(R). Then there is a regular element dE cJ(R). with I g di(R) g I, . Thus, again we have I. -- dR, which implies dE r-st(R). On thg other hand, since dE cJ(R), we have dR c cJ(R), because dR is a v-ideal and cJ(R) is. notav-ideal. Thus I = dR ( cJ(R) c I , acontradiction. HenceI= cJ(R).. v -v. Remark. In the case 2 is fmite dimensional over its center, [Di] has obtained the following ([MMU, (7.l2) and (7.5)]): (1) 0, (I) = O, (I) for any R-ideal L. (2) If cR 2 Rc for some cE 2, then cR = Rc. In particular, r-st(R) = st(R) = l-st(R). {cE2 lcR is left stabilizing }. (3) If cR = Rc, then cS = Sc for an overring SofR.. However, if 2 is inf/mite dimensional over its center, then (1) - (3) are not necessarily to be held. For example, let (K,M and (K, 7) be valued fields as in [XKM,. (2.5)], namely, lfD V, 6 E Aut (K) such that 6(V) = Vand 6(M c YV. Set S = M(i)=. '. ' 15. '. '.

(19) M+ xT and R = V(,)= V+ xT, where T= K[x,o]<.), the localization of K[x, 6] at maximal ideal (x) =, xK[x, 6]. S and R are both Dubrovin valuation rings, in fact, they. are total valuation rings. First we note that xSx-` = 6 (S) = 6(M + 6(xT) g if+ xT = S. By [XKM, (1.5)], 6 (S) c S, so that xSxr' c S. HenceI= Sx is an ideal ofSwithI ). xS. Similarly, xR = Rx, because 6(V) = V. Furthermore, it is easily seen that S=. ' = O, (i) . Hence (1)--(3) are' not necessarily true. In particular, xE l- st(S) O, (D c• x-' Sx butxÅë st(S).. '. 2.2. R-ideals ofa Dubrovin valuation ringR For any ideal I of a Dubrovin valuation ring R of 2, we write Spec(R) for the set. ' and denote .v7-= A{Pl P E Spec(R) with P 2 I} theprime of all prime ideals ofR radical ofI which is a prime ideal ([MMU, (13.1)]). An idealA of R is called right. V-Z--primary if aRb gA, where a, bER, implies either aE A or bEe. Similarly, left primary ideals are defined. In [BMU], they have described all right primary ideals of R. So it is natural to ask the question: Describe the structure of all R-ideals by usage of stabilizing elements and primary ideals. In this section, we give a partial answer to this. question in general case and a complete answer in the case 2 is finite dimensional over its center after a few prelinimary lemmas.. '. Lemma 2.2.1. LetR be a Dubrovin valuation ring and letI be an R-ideal which is not. finitely generatedasarightS-ideal, where S= O.(D•Then (S:I)i = (R:I)i•. Proof: First note that J(S) =,J(S)2 and so I = I .r(S), by Lemma 1.2.8 and Proposition. 1.2.10(3). Hence we have (R:I), = (R :U(S)),= ((R:J(S))i :I)i = (0i(J(S):I)i = (s :D, by [MMU, (6.8)], because (R :J(S)), = O, (J(S)) •. Remark. In Lemma 2.2.1, we can not drop the condition that I is not fmitely generated as a right S-module and J(S) =J(s)2 .. 16.

(20) (l) IfS=R and I= aS (see the example in the Remark of Sec. 2.1), then (R :I), g R a'i g Sa-i = (S :I), •. '. ' so (R :I), g R a-' g Sa-i = (S :I), , (2)' If J( S) DJ(S)2 , then J(S) = aS, and. . wherel=aS. A Goldie prime ideal P is Archimedean if there is a prime segment P DP, which is. Archimedean, that is, for any a G PXP,, there is an ideal IgP with a E I and ,P,=. AJ" (see [BMO] and IBMU] for details concerning prime segments). Then we have the following.. Lemma 22.2. (1) Suppose that J(R) is Archimedean andJ(R) :RxR. Then J(R) isprincipal.. '. (2) LetIbe an R-ideal with S= O,(I) and T == O,(I). Suppose thatI = RqR and J(E}) isArchimedean. ThenI = aS == Tafor some a E I.. Progif] (1) Let J(R) D P, be the Archimedean prime segment. But gr = {A ]A is an ideal. ofR and A ) x}. Then gr is a non-empty inductive set, and so it contains a maximal element B. Since there are no ideals between J(R) and B properly, B is prime if J(R) =J(R)2. In this case, we have B = P, , which contradicts the Archimedeaness. So J(R). DJ(R)2 and hence J(R) is principal. ' (2) To showI= aS for some a E I, it suffices to prove that IJ(S) c L Suppose that on the contrary, I= ILI(S). Then q= ri gxi+ ... + r. qx. and Sx = Sxi+ ••• + Sx., Where r, E R and x, x, E ..r(S), i= 1, ..., n. Now, we haveI=ISxS. If .J(S) = StS, then by (1), .r(S) = sS = Ss and so I = Is. It follows that s-` E O. (I) = S, a contradiction. Hence .J(S). ) SxS and there is some t E J(S) with J(S)tD SxS by Lemma 1.2.4. Then I= IStS g IJ(S)t g It and so t-i E0, (I) = S, a contradiction. Hence ID IJ(S). Thus I = aS for. some a E I and I " aSa-ia= O, (I)a= Ta follows.. 17.

(21) Theorem 2.2.3. Let R be a Dubrovin valuation ring in simp le Artinian ring 2. Let I be. an R-ideal such that O, (I)= S = 0, (I) andI is notfinitely generated as a right R-. ideaL Suppose that J(E}) isArchimedean. ThenI= cAfor some cE st(1sp andA a right and left J(S)-primai v ideal•. ProqC Let J(S) D P be an Archimedean ptme segment. By Lemma 2.1.2 (2), we have (.R :I),I = .I(S)• Let gn = {xE(R:I), ] tw= J(S)}. Then gn = SZ5, because J(,S) is Archimedean. First we claim thatA = SbeSI (xE gr ) is right and left J(S)-priniary. Since. VI4-= J(S), it suffices to prove that O, (A) =S=•O, (A) by [BMU, (2.5)]. It is clear. '. '. that O, (A)R S and O, (A)p S. If O, (A)) S, then O, (A)2 R, because there are no Goldie prime ideals betvveen J(S) and P, and' so we haveA = BR. =R. , a contradiction. Similarly, O, (A) = S.. '. Next we show that U{ ,SbeS lx E gr } = (R:I),• Since O,(I)=S= Oi(I),. '. (R :I), is a right S-ideal. To show that (R :I), is a left S-ideal, frrst suppose that I= aS. for some aE L Then, by Lemma 2.1.1, (R :I), =' J(S) a-' so that it is a left S-ideaL. '. Suppose that I is not fmitely generated as a right S-ideal. Then, by Lemma 2.2.1, (R :I), is a left S-ideal. Hence ,SbeS g(R :I), for any xE 7 . Suppose that y E (R :I),. but y e gn . Then we have Sy SI g P and so SyS g SxS for any x E gn . This is a contradiction and hence U{ SxS Ix E gn }= (R :I), holds•. '. '. '. Finally we claim that' 0, (,SXS)= S for some x E gn . Suppose that O. (,SbeS)D S for allx E gr . Then O, (,SboS)2 R, and so (R :I),' R. = (R :I)i•. Case 1. I= aS for some aE L Then S" O, (I) = aS a-i and J(S) = a.J(S) a-i. foilows. By Lemma 2.1.1 (1), (R:I), =(R:aS), = (R:S),aMi = .J(,S) a"= a'iJ(S), and so (R :I), = (R :I), R. = a-i J(S) R. = a'i R.. Hence we have J(S) = R. , a '. contradiction.. 18.

(22) Case 2. I is not fmitely generated as a right S-ideal. Then, by Lemma 2.2.1, I 'i = (R :I), and so I -i R. = I 'i . It fo llows that R. g O, '(I -i) = O, (I) = S by [MMU, (6.10)], a contradiction. '. Hence there is some xE Y such that O (SbeS)= S. The above discussion shows. r. that there exists xE gr such that A = ESbeSI is right and left ,J(S)-prirnary, where O, (iSbeS)= S = O, (iSbeS). By Lemma 2.2.2, there is some cE ,SboS such that StcS ; cS = '. Sc . Thus we have cE st(S) and I= c-iA (e-i E st(s)).. '. '. A Goldie prkne ideal P is called a limit prime ideal ifP = U{P, 1P D PA : Goldie ptme}. Suppose that 2 is finite dirnensional over its center. Then any prime ideal is Goldie prime and it is either Archimedean or limit ptme <see IBMO]). Also note that an ideal is right primary if and only if it is Ieft primary, which is called a primary. ideal (see [MMU, (13.4)]). Now we have the following proposition which describes all. R-ideals in terms of pimary ideals and stabilizing elements in the case 2 is fMite dimensional over its center.. Proposition 2.2.4. Let R be a Dubrovin valuation ring of a simp le Artinian ring 2 itvith finite dimension over its center andI be an R-ideal with O. (I) == S ( = O, (J)? . Then. (1) Suppose that I isfinitely generated as a right R-ideal. Then J = cR == Rcfor some. cE st(R). (2) Suppose that I is notfinitely generated as a right R-•ideal.. (a) lr J(S) isArchimedean, thenI= eA =Acfor some cE st(S) and some J(&primary idealA. (b) IfJ(S) is limitprime, thenIis one of thefollowing three;I= cS =Scfor some eE st(S),I= cJ(R) = J(R)cfor some cE st(R) andI" nc,R,for some c2 E st(RP, where RA = Rp, and PA runs over allArchimedeanprime ideals with P, c J(,S).. 19.

(23) PToof (1) Because R is Bezout, we have l= cR for some cE st(R) and so I= Rc by Rern-ark in Sec. 2.1.. (2)(a) This follows from Theorem 2.2.3 and [MMU, (7.1 1)]. (b) First we shall prove that J(S) = U{P,:?trchimedeanI P, c .1(S)}. To prove this,. let .r be any non-zero element in J(S) andA = S5cS. Suppose that O, (A) = T. ThenA. =yT = Ty by [MMU, (7.10)]. ThusP=fi P, = fiA" is an Archimedean segment (see [BMO, (5)]) and x E P.. Case 1. IcI.Then J(R) =J(R)2 and I= cJ(R) for some cE st(R) by v ' 1 .3. Proposition 2. ' Case 2. I= i, . Suppose thatI# cS for any cE st(S). Then, by Lemma 1.2.8,I=. IJ(S) = I( U, P, ). If I = IP, for some X, then S = O, (I) 2 O, (Pi) = Rp, ,a contradiction. So we have IRi =) J =) IPA. To show that 0,(IRz)= Rz, suppose. that O,(IR,)= T )Rz. Then IR,= IT 'and so IT = ITP,=IR, P, 'IPa c I, a contradiction. Hence O. (IRP = R, . So, by the similar method as in Lemma 1.2.8, we. have IR,= cAR,, for some c, EIR,, because IRz=R2Iby [MMU, (6.5) and (7.11)].. Hence IRi' c2R2= R,cz by [MMU, (7.5)]. Thus cA Est(R, ). To show that I= fi IR,,let B= n IR,.Then (R :I),B( (R :I), IR,=J(S) R,= R, for any X by Lemma. 2.12 (2). So (R :J), B g n R, = S by [BMO, (4)]. Thus B g (S : (R :l),). = (S : (S :I),),= I,=Iby Lemma 2.2.1 and Proposition 1.2.10, and henceI= n c,R,for some c2 G St( RA )•. '. Remark Proposition 2.2.4 (2)(a) is not necessarily held if 2 is inf/inite dimensional over. its center. To give a counter example, let S = M+ xT and R = V+ xT be the same as in. the example of Remark in Sec. 2.1 and setI= Sx. Then O, (I)=S andIis not fmitely generated as a left R-ideal. Assume that I =Ac for some ce st(S) and some P-prirnary. idealA, where P = J(S). Then, by [XKM, (1.10)(3)], we may assume that for some cG. st(rv) = KX{O}. By Remark to [XKM, (1.7)],A = 2 + xT for some non-zero priinary. .v A = Q(A), where rp: T=K[x,a](.) -> Kis the natural map with rp ff(x)c(x)-i) = ideal. ' 20.

(24) -1(rtx) fo Co. = fo +fix+•••+f.xnand c(x)=co +c,x+...+c.xM. follows that O = xS A K= cA A K= cA"'. , a contradiction.. 21. with c,#O). So it.

(25) CHAPTER 3 A characterization of fully bounded Dubrovin valuation rings. A ring is called right aeft) bounded if any essential right (left) ideal contains a non-zero (two-sided) ideal. A ring isjust called bounded if it is both right bounded and. left bounded. Let Sbe aring. We say that Sisfully bounded if S/P is bounded for any prime ideal P of8 We write J(,S) for the Jacobson radical ofS and Spec(S) for the set of all prime ideals of S.. Let R be a Dubrovin valuation ring in a simple Artinian ring 2 (see [MMU, Chap.II] for the definition and elementary properties of Dubrovin valuation rings). A. prime ideal P ofR is called Goldieprime if R/P is a prime Goldie ring. We denote by. G-Spec(R) the set of all Goldie prime ideals of R. Now letP,,P E G-Spec(R) with Pi D P• The pair P, D P is called aprime segment if there are no Goldie primes properly. between Pi and P.. Let P E G-Spec(R) with P # ,J(R) and setP,= A{P,I P, E G-Spec(R) with Pz D P}. In [BMO, (6)], they have shown that the following four cases only occur: (1) P is lower limit, i.e.,P=P, . Otherwise, P, DP is aprillle segment.. (2) P, DP is Archimedean. (3) P, DP is simple. (4) P, DP is exceptional, i.e., there exists a non-Goldie prime ideal C such that P, ). CDP. In section 3.1, we prove that R is fu11y bounded iff (1) and (2) only hold (see Theorem 3.1.5). (Note that RIJ(R) is bounded, because it is a simp}e Artinian ring). For. any regular element c in ,J(R), we define P(c) =A{ PA1 PA E G-Spec(R) with cE P2 }, a Goldie prime ideal ([BMO, (1)]). R is called locally invariant if eP(c) = P(c)c for any regular element c in J(R). This concept was defmed by Grater [G] in order to study the approximation theorem in the case where R is a total valuation ring. We show that R is. fu11y bounded if and only if it is locally invariant, by using Theorem 3.1.5 (see Proposition 3. 1 .6).. 22.

(26) In section 3.2, we give several examples of of fully bounded Dubrovin valuation rings of e with infinite dimension over its center. If e is of fmite dimensional over its center, then R is always fu11y bounded.. 3.1. Fully bounded Dubrovin valuation rings Throughout this section, R will denote a Dubrovin valuation ring in a simple. Artinian ring 2. For anyP E Spec(R), set C(P) = {cERlc is regular mod P}. IfP E G-Spec(R), then C(P) is localizable and we denote byR. the localization of R at P.. Before starting the lemmas, we note the following: there is a one-to-one correspondence bgtween G-Spec(R) and the set of all overrings ofR, which is given by. P -> R. with P = J( R.) and S -> 1(S) (P E G-Spec(R) and S is an ovening of R).. '. Furthermore, for anyP, , P E G-Spec(R), P, DP iffR.cR.,([MMU, (g 6)] and [BMO, (g 2)]). We will use these properties throughout the chapter.. Lernma 3.1.1. Let S be an order in 2 andA be an S-ideal such that O,(A)" T". O,(A)where O.(A)" {qEe1AggA} and O,(A) = {gE2IgA gA }. Suppose that A " aTfor some a E A• Then A = Ta.. Pro of T = 0, (A ) = aTa"i imp lies A = Ta.. '. '. '. Lemma 3.1.2. Let R be a Dubrovin valuation ring of 2 and PE G-Spec(R). Suppose. that P is lower limit, i.e., P= A{P,I P,E G-Spec(R) withP, ) P}. ThenR.= wR., and C(P) = uCp, •. 1?roof: Since PADP, it follows that R. )R., so that R. 2S =uR., . Suppose that R. D S. Then for any P, ,P, = J( R., ) p J(S) D ,J( R,) =P implies P =A P, p .J(S) D. P, acontradiction. HenceR. = uRp, and so C(P) = vC(Pz) follows.. 23.

(27) Lemma 3.1.3. Let R be a Dubrovin valuation ring of e andP E G-Spec(R). Then (1) Spec( R.)={P, IP, G Spec(R) with P 2 Pi }•. (2) Let P,andP,be in Spec(R) withP2P, DP,• ThenPi DP, is aprime segment of R f and only f it is ap rime segment ofR. •. Proof (1) Let P, E Spec( R. ). Case 1. If P, is Goldie prime, then (R. )., is an overring of R. (and so of R) with 1( (R. )., )= P, , i.e., P, e Spec(R) andP= J( R. )2P, . Case 2. If Pi is non-Goldie prime, then we can construct an exceptional prime. segment ofR. , say P, DP, DP, by [BMO, (6)]. By case 1, P pP, and P, , P, E G-Spec(R). It easily follows from note before Lemma 3.1.1 that there are no Goldie primes properly between P, and P, , which implies P, DP, is a prime segment ofR. As. in [BMO], let K(P,)= {aE P,1 P,aP,cP, }. Then K( P,)= P, by [BMO, (7)] and so P, D P, is an exceptional prime segment of R with K( P, ) = P, , i.e., P, is non-Goldie prime of R with P DP, . Conversely, letP, e Spec(R) with P ;P, . Then from note before Lemma 3.1.1 and the method we have just done, we can easily see thatP, E Spec(R.)and that P, E G-Spec(R) iff P, E G-Spec(R.). (2) This is clear from (1).. Lemma 3.1.4. Let R be a Dubrovin valuation ring of 2 and P, )P be an Archimedeanprime segment. Thenfor aay cE P, XP, thefollowing hold:. (1) Rp,cRp,"aRp,=Rp,a,forsomeaEPi• (2) ,if c is a regular element, then cRp, = Rp, c and cPi " Pi c•. Proof Firstiy note that P, D P is an Archimedean prime segment ofRp, by Lemma 3.1.3 and [BMO, (7)]. (1) Let RN .,= R.,/P, a Dubrovin valuation ring of R. = R.IP (see Theorem 1.2.6). -N -v -. such that J(R.,)= P,= P, /P andP, )(O)is Archimedean. Here for anyaE R,,,we. '.

(28) write a". -VR.,.-VIf P,=P, At -V2, then At -V for the image ofa in O# N R., c". Rp,= a- Rp,=. R.,afor some ae P, by [BMU, (2.1)]. If P,DP, 2, then R.,is a Noetherian Dubrovin valuation ring and so any ideal of fip,is power ofP-' a'". i• Thus R"'. p, c"' R-'. p,". Rp,= Rp, aN for some a E Pi, becausePi is pimcipal. Hence, in both cases,. Rp, 'c Rp, + P = aRp, + P =Rp, a + P• However, since a E Cfi,, (6) = { b-'. ' regular in R'V. E ,iE,, l b'-. t.. ., }, it follows that a E C.,, (P)and so a is a regular element by. .Proposition 1.3.2. Thus we have aRp, a-ig aRp a-i= Rp. It follows that aRp, and P. are both left aR., a-iand right R.,-ideals. Hence aR.,D P by [MMU, (6.4)] and similarlyR.,a =) P. Since R.,cR., and P are both ideals of R.,, it follows that Rp, cRp, DP. Therefore Rp, cRp, = aRp, =Rp, a follows.. (2) By (1), Pi RR.,cRp,= R.,a=aR., for some aE.P,. Suppose that cR.,c Rp,cRp,.Then, by Lemma 1.2.4, there is ab ERp,cRp, such that cRp, g bPi g aPi, because 9 (c Rp,) = cRp, c-i and Pi = J( Rp, ). So Rp, a-icRp, g Pi. On the other. hand, Rp,cRp,= aRalmplies that Rp, a-icRp,= Rer a contradiction. Hence, cRp,= Rp,cRp, and similarly Rp,c= Rp,cRp, so that cRp,=Rp, c. Since cRp, c-i== Rp, and 1( Rp,' )= Pi, we have cPi c-i= Pi and so cPi = Pi c.. ' Theorem 3.1.5. Let R be a Dubrovin valuation'ring of a simple Artinian ring 2. Then R isfully boundedf and only ffor aay P E Spec(R), P # J(R), thefollowing hold:. (1) PEG-Spec(R). '. (2) P is either lower limit or there is aP,E Spec(R) such thatPi ) P is an. Archimedeanprime segment`. '. Proof Suppose that R is fu11y bounded.. (1) Assume that there is a non-Goldie prime ideal C Then we have an exceptional. prime segment, say, P, DCD P, by [BMO, (6)].Ris an n-chain ring by Theorem. 25. js.

(29) 1.2.2 and so is R= R/C. This implies that Rhas a finite Goldie dimension, say, m (S n). Thus there are non-zero uniform right ideals U, of R such that U, e ... e U. is. an essential right ideal of R.Since R isaprime ring, U,AP,2 U, P,# Oand so. '. there are non-zero il;. E l7;. A ]Pi, where u, E Pi.Set I= uiR + .. .+ u. R. ThenI= aR. for some aE I, because R is Bezout (Theorem 1.2.2) and I = u, RO ... e u. R= a R is an essential right ideal ofR. We claim that P, DI . On the contrary, suppose. that P, =I, i.e., P,= aR +C Note that O, (C) = R.,= O,(C) by [BMU, (2.2)] so that C is an ideal of R.,.If C is aprincipal right ideal of R.,,say, C=cR., for some cE. C, then Pi=aRp,+cRp, =bRp, for some bE Pi. It follows from Lemma 3.,1.1 that. ' Pi2=) C, which contradicts to the fact that there are no Pi=bRp,= Rp,b and so PiD ideals properly between P, and C (cf. [BMO, (6)]). If C is not a principal right ideal. of Rer theR CP,=Cby Lemma 1.2.8 and so P, = P,2=aPi +CPi= aPi +C Thus we have a= ap +dfor somepE P, and dE C and a(1-p) = dE C It follows that aE' C, because 1-p isaunit of Rp,,which shows I= O,acontradiction. We have. shown that P, =)Iand Iis an essential right ideal of R.Hence R is not bounded, because there are no ideals properly between P, and C Therefore, any prime ideal of R is Goldie prime.. (2) Let P E G-Spec(R) and suppose that P is not lower limit. Then there is a P, E. G-Spec(R) such that P,D P is a prime segment, which is not exceptional by (1). Suppose that this is simple. For any cE P, A C(P), it follows that c P, is an essential. right ideal of R = RIP, which is a Dubrovin valuation ring of R. IP (Corollary 1.2.3).. Suppose that cP,= Pi,i.e.,cP,+P= P,.SincecP, and P are both left cRp, c'i and right Rp, -ideals (note cRp, c-ig cRp c"i = Rp ), we have either cPiDP or cPigP by Proposition 1.2.5. The latter case is impossible and so cP, D P. Thus cP, = P and c-i e Ot (Pi ) = Rp, follows. This is contradiction, because cE Pi . Hence we have shown. that P,D cP, and cP, is an essential right ideal. Therefore, R is Bot bounded,. 26.

(30) because there are no ideals properly P, and (O). Hence either P is lower 1imit or there. is a P, e G-Spec(R) such that P, DP is an Archimedean prime segment. Conversely, suppose that the conditions (1) and (2) hold and let P E Spec(R). Then P is Goldie prime by (1). Firstly, assume that P is Iower limit, i.e., P = A{ P,I. P, E G-Spec(R) with P, )P}. Then C(P) =v C(P, ) by Lemma 3.1.2. So, for any ce C(P), we have cE C( Pz)for some X. Then cR )Pz,because cR and Pz are both left. cRc-' and right R-ideals. Hence c R=)Pz # Oin R= RIP, showing thatRis bounded.. '. Secondly, suppose that the prirne segment P, = P is Archimedean and let cE C(P). Then, as before, c P, is an essential right ideal of R=RIP and so cP, A C(P) # SZS. Let. de cp, A C(P). Then, by Lemma 3.1.4 (2) and Theorem 1.3.3, cR p cPp dR., =R., d and dR., D P follows. Therefore, R = RIP is bounded and hence R is fully bounded.. As an application of Theorem 3.1.5, we have the following:. Proposition 3.1.6. Let R be a Dubrovin valuation ring of a simple Artinian ring 9 Then R is locally tnvariantf andonly f it isfully bounded.. Proof. Suppose that R is locally invariant. In order to prove that it is fully bounded, on. the contrary, assume that R is not fully bounded. Then there are prime ideals P, P,. such that either the prime segment P, D P is simple or P,E G-Spec(R), P is a non-Goldie prime ideal and there are no ideals properly between P, and P. In either case, we shall prove that there is a regular element cE PiXP. Let ci be any element in P, XP. If c,R is an essential right ideal, thenc = c, is regular. If c,R is not an essential. right ideaL then there is a right ideal I such that cR e Iis essential. So it follows from. Goldie's theorem that (cR e DP, is also an essential right ideal which is contained in. P, but not in P. So there is aregular element cE (c,ReD P, but not inP by [MR, (3.3.7)]. Now let cE P,XP such that c is regular. Then cP,= P, c, becauseP,= P(c).. 27.

(31) SinceP, ?cP,= P, cD P, we have cP, =P, , which impliesc-' E Oi(P,)= R., . Hence Rp, = cRp, gPi , a contradiction. Therefore, R is fully bounded.. Suppose that R is fu11y bounded. Let cE .J(R) such that c is regular. By the. assumption and Theorem 3.1.5, P(c) = A{ P,l P, E Spec(R) such that P, ) c}, which is Goldie prime by Proposition 1.2.13. Suppose that P(c) is upper 1imit, i.e., P(c) =. u{P, P,E G-Spec(R) such that P,c P(c)}. Then there is aP,withPp c. This contradicts the choice of P(c). Hence P(c) D P = u{P" P(c) D P, } is a prime segment which must be Archimedean by Theorem 3.1.5. Since cE P(c)XP and c is regular, we have cP(c) = P(c) c by Lemma 3.1.4. Hence R is locally invariant.. We say that R is invariant if cR c-i = R for any regular element c in R and that it is of rank n if there are exactly n Goldie prime ideals. From Lemma 3.1.4, we have. Proposition 3.1.7. Suppose that R is Archimedean and is of rank one. Then it is lnvarlant.. Proof. Let c be any regular element and let c, be any regular element in J(R). Then we. have cRc-i=cciR(cci)-i=R by Lemma 3.1.4, because ci , cc, E J(R)•. 3.2. Examples We will give several examples of fu11y bounded Dubrovin valuation rings.. ' Example 3.2.1. Any Dubrovin valuation ring of a simple Artmian ring with finite dimension over its center is fuIIy bounded.. Example 3.2.2. Any invariant valuation ring of a division ring is fu11y bounded (see. [XKM, (Remarks to Examples 2.1 and 2.4)] for invariant valuation rings of division rings with infinite dimensions over its centers).. 28.

(32) In order to give more general examples, we recal1 the skew polynomial ring 2[x,6] over 2 in an indeterminate x, where 6 E Aut(2). Since 2[x,6] is a principal ideal ring, the maximal ideal P = x2[x,6] is localizable, i.e., T = e[x,o].= ff(x) c(x)M' 1 f(sc) G 2[x,6] and c(x) E C(P)}, the localization of 2[x,a] at P, is a Noetherian. Dubrovin valuation ring withJ(7r) =xT. Since 2 is a simple Artinian ring, C(P) = {c(x). e e[x,6] l c(x) = c,+c,x+...+c.x" such that cois a unit in 2}. For any t =f(x) c(x)-' E T, wheref(x) = f, +f,x+...+f,x' and c(x) = c, +c,x+...+c.x", the map q: T -> 2 defined by Q (t) = f, c,-'is a ring epimorphism. Now let R be a Dubrovin valuation ring of 2. Then, by [XKM, (1.6)], R = tp'i (R), the complete inverse image ofR by Q, is a Dubrovin valuation ring of 2(x,6) (9(x,o) stands for the quotient ring of. 2[x,6]). Furthermore, let P = sD iE(sD E Spec(R)). Then P E Spec(R-) and RN IP i R18Z) by [XKM, (1.6)] and its proof. Thus it follows from [XKM, (1.6)] that R-' fu11y bounded iffR is fu11y bounded. Hence we have. Example 3.2.3. With notation above, suppose that R is a fu11y bounded Dubrovin valuation ring of 2 and that 6 is of infinite order ([XKM, (Examples 2.1 - 2.6, 2.7 and. 2.8)]). Then fi is a fu11y bounded Dubrovin valuation ring of 9(x,6) and e(x,6) is of infinite dimensional over the center.. Finally, we give a few remarks on non-fu11y bounded total valuation rings: An example of a total valuation ring with a simple segment was first constructed by [Mt].. See [BT] for other examples of total valuation rings with simple segments. Dubrovin. constructed an example of a total valuation ring with an exceptional prime segment ([D3])•. 29. is.

(33) CHAPTER4 ' Non-commutative v-Bezout rings. Throughout this chapter, Vwill be a total valuation ring of a division ring K,. i.e., for any nonzero kE K, either kE J7 or k-i E V. Let Qobe the semigroup of. nonnegative rational numbers and 6 be a semigroup homomorphism from Qoto Aut(pt), the group of automorphism of V, i.e., 6 (r + s) = c(r).6(s) for any r,s E Qo•. Furthermore, R = l7[x",o1rE Q,] isaskew semigroup ring of Q, over JZ, i.e., it is a. ring with left V-basis {x"lr E Q,}. Each element of R is uniquely a finite sum aix"' +...+akx"kwith a, E V. The multiplication is defined by x"a= 6(r)(a)x"for any a. E V and r E Q,. Since 6 is naturally extended to a semigroup homomorphism from Q, to Aut(K), we have T= K[x',6Ir E Q,] is a skew semigroup ring ofQ,over K. In Section 1, we prove that R = V[xr,o l r E Q,] i's v-Bezout, which is defined. in [Ma] and is a non-commutative version of commutative GCD-domains. In Section 2, we give some examples of non--commutative v-Bezout rings with some types of automorphisms.. 4.1. Non- commutative v-Bezout rings Let S be an Ore domain with its quotient ring 2 and let I(J) be a right (left) S-. ideal. We use the following notation [MMU]: (S: D, = {g E 2l glg S}, (S: J) , = {g E 2lJg g S}, I, = (S: (S: D,), and .J= (S: (S: .1),),. It is clear that I. (.J) is a right (left) S-ideal containing I(D, respectively. IfI = I. (J = .J), then it is called a right. aeft? v-ideal. An Ore domain S is called right v-Bezout if I. is a principal for any finitely generated right ideal I of S. Similarly, we can define left v-Bezout and S is said to be v-Bezout if it is right v--Bezout as well as left v-Bezout.. A partially ordered set A with ordering 2 is called an ascending net if for any. Zi, Z2 in A, there is a X E Awith Z,S X (i -- 1, 2). Then we have the following. lemma.. 30.

(34) Lemma 4.1.1. Let A be an ascending net and letRz be an Ore domain with its guotient division ring KA,for each X E A. Suppose thatR, gR2 ifp SX. SetR=. u{R, 1XE A}andK=u{K, 1XE A}. Then (1) K is a quotient ring of R which ls a division ring.. (2) IfR, is aBezout ringy'or all X E A, then so is R.. ' (3) LetP, be a completelyprime ideal of R,, which is localizablefor aay X E A. Suppose that P,AR," P, i:leX2p. Then (a) P= w{ P, lX E A. } is a completelyprime ideal ofR and is localizable.. (b) Rp=W{Rz,, lXEA}' (C) IfR2,, is a total valuation ringyCor all X E A, then so is R... Let N be the set of natural numbers. Then it is considered an ascending net in. i. the following obvious way: n2m iff mln for any m, n E N. Let R.= V[xS, 6] = {akxff+ ••• + aixff+ ao l a,EJ7 }. Then R. is considered as a skew polynomial ring. '. l! 1!. over J7 in the indeterminate x" with x"a = 6(-. il)(a)xn for any a EV• Let P." ,J(V). .1 .. [xi, 6], a cornpletely prime ideal ofR.and it is localizable such that R.,. is a total. i. valuation ring of K(xff, 6) (see [BT]). Obviously, R.2 R. and P. = P.AR. if n 2 m. Furthermore, P= J( P')[x', 6l r E Q,] = u{ P.l n E N }. Let R = V[ x', d r E Q,],. i. T= K[x", 61r e Q,] and let T.= K[xV, o], be a principal ideal ring for each n E N.. ' Then it is obvious that R = U..o R, and T= U..o T. . So from Lemma 4.1.1, we. '. have the following:. 31.

(35) Proposition 4.1.2. (1) P = J(P')[x', c ir E Q,] is localizable andR. is a total valuation ring with Rp =U Rn,. '. (2) T= K[xr, o1r G Q,] is aBezout ring with its guotient ring K(x', 61r E Q,)•. ' Let 6 be a left T-derivation of V, where T E Aut(J7) and assume that (:, 6) is compatible, i.e., 5(J(") g .I(V). Let S= J7[x; T, 6] be an Ore extension over Vin an indeterminatex. Then P :.J( V)[x; T, 6] is localizable and S.,the localization ofS at P, is a total valuation img (cf. [BT]). Now letf(x), g(x) G S and let I = Si7(x) + Sg(x). Then. S.I= S. a, for some a E l7 and K[x; T, 6] I= K[x; T, 6] b(x), for some b(x) EK[x; :, 6].. There are b E K and b,(x) E S XP with b(x)a-i= bb,(x). With these notations, we have the following:. Lemma 4.1.3. [Ma, (2.1) and (2.3)]. .I= S.IA K[x; T, 6] I= Sc(x), where c(x) = b, (x) a E 8. '. '. By using Lemma 4.1.3, we have the following theorem which is inspired by [C, (35)].. Theorem 4.1.4. Let Vbe a total valuation ring of a division ring K. Then R = l7[x", 6. l r E Qo] is v-Bezout, and it is not Bezoutf V#K. Proof Let I = RLKx) + Rg(x), for somef(x), g(x) e R. There is a natural nurnber m such. thatf(x), g(x) e R. . Set I.= R,f(x) + R.g(x). Then by Lemma 4.1.3, there is c(x) E. R.with R.,. I.AT. I.= R. c(x), where R.,. I. = R.,. a (aE to, T. I.= T.b(x) (b(x) E T. ), b(x)a"= bb, (x) (bE K, b, (x) E R.X P.) and c(x) = b, (x) a. For any natural number n with m1n, we have R.,. I.= R.,. a and T. I.= T. b(x) aiid SO ,In= R.,. I.A T. I.= R.c(x) bY Lemma 4•1•3• Sincef(x), g(x) E R. c(x) gRc(x), it follows. 32.

(36) ' '1. that I Åí. -R.c(x.),. Suppos• e that Ir,= Ra for s• ome oc E .k.r(x', s!r c- Q,). Then or E .k.{(x"n ,. 6), the quotient img of T., for some n and we assume that mln. It follows that I. g. 1 RaA K( xn , 6) = R ct and so R c(x) = I g R ec Thus Rc(x) g Ra and hence I=. mn. nv. vn. Re(x) follows. IfJ= J,+ J,, where J, and J, are Ieft ideals of R, then it is easy to. Check that .J= ,(,J, + J2)=,(,Ji + ,J,) and so R is left v-Bezout by induction on generators. Similarly, R is right v-Bezout.. Now, suppose that R = V[x", 61r E Q,] is left Bezout. Let ct be a non-unit element in VX{O}. Then there exists h(x) E R such that Ror + Rx = Rh(x). We have ct = a(x) h(x) and x = b(x) h(x) for some a(x), b(x) E R. Then it follows that h(sc) is constant,. say, h(x) = c and b(x) = b,x for some b, E M Thus 1 = b, 6(1)(c) and so c is unit in JZ.. Then Ra+ Rx = Rh(x) = Rc = R implies that ais unit in V, a contradiction. Hence R is not left Bezout.. 4.2. Examples Finally, we wil1 give several examples of skew semigroup ring of Q, over total valuation rings.. Example 4.2.1 (trivial case, 6= 1). R = l7[x' lr E Q,] is v-Bezout, where Vis any total valuation ring.. In order to provide non-trivial examples, let K = F({ Y, }l t E Q) be the rational. function field over a field F in indeterminates { Y,lt E Q}, where Q is the field of rationals. For any r E Q,, let 6, E Aut(K) determined by; cr, (a) = a for any a E F and. 6,(Y,)= Y,., for anytE Q• Furthermore, let v be the valuation of K detemined by. v(a) =O for all aEF and v( Y,)=1 for all tE Q. Then JZ= {k EK1 v(k) 2 O} is a discrete rank one valuation ring of K. It is easy to see that 6,(to = Vfor any r E Q,. 33.

(37) and o,., = a,.o, . Hence the mapping 6 : Q, -> Aut(V) defined by 6 (r) = a, for. any r E Qo is a semigroup homomorphism.. Example 4.2.2. With 'the notation and assumption the above, R = J7[x" l r E Q,l is vBezout which is not Bezout.. '. '. In order to get another example which is not discrete rank one valuation ring, let G = e Z, (r E Q, Z,= Z), the direct sum of the copies Z, which is a totally ordered. abelian group by lexicographic ordering and let K and o,be as in Example 4.2.2. We define a valuation ofKas follows: v(a) =O for all a E F and v( Y,) = (..., O, 1, O, ...) E. G, the t-th component is 1 and the other components are all zero. Then V= {k E Kl v(k) ) O} is a valuation ring ofK with infinite rank and J( V) = J(J7)2. It is not hard to. see that 6, (" = JZ for all r E Qo. Hence, we have. Example 4.2.3. R = J7[x' l r E Q,] is v-Bezout, where Vis commutative valuation ring with infinite rank and J( M = J( v) 2 .. In order to give an example of non-commutative valuation rings, let be Vo be any total valuation ring of a division ring K, and G = <g, l r E Q> be a group which is isomorphic to Q, i.e., g, .g, = g,., for any r, s E Q. Since Gis abelian, the group ring. V, [G] and K,[(n have the same quotient ring K,(G) which is a division ring. As before, for any r E Q,we defme an automorphisma, of K, (G) as follows: o, (a) = a for aH a E K, and q(g,) = g,., for any t E Q. Now J( V, )[G] is localizable and V= V,[G],(.,)[.] is a total valuation ring of K, (G) (see [BMO, (2.6)]). Since 6. (J(E)[G]) =. J(l7)[G], 6, is considered as an automorphism of Vwith o,., = a..o, for any r, s E. Q,. So the mapping 6:Q, -> Aut(J7) given by 6 (r) = e, for anyrE Q, is a semigroup homomorphism.. 34.

(38) Example 4.2.4. With the notation and assumption the above, R = Bezout but not Bezout.. 35. nxr lrE. Qo] is v--.

(39) CHAPTER5 Overrings of Non-commutative PrUfer rings satisfying a polynomial identity. In [AD], they defined the concept of non-commutative PrUfer rings in the context of prime Goldie rings and studied the structure of Priifer rings. In the case when prime rings satisfying a polynomial identity (PI), Morandi studied PI Pr{ifer rings. under some conditions such as; integral over its center or the center is commutative PrUfer. Furthermore, Dubrovin [D2] proved that any prime ideal of a PI PrUfer ring is localizable.. In Section 1, we describe the properties of ovenings of PI PrUfer rings.. In Section 2, we describe prime ideals of any overring of a PI PrUfer ring by using some results in [D2] and [Mo]]. We refer the readers to [MMU] for elementary properties of Pnifer rings and Dubrovin valuation rings.. 5.1. 0verrings efPI Pnifer rings Throughout this chapter, R will be a prime Goldie ring with its quotient ring 2. LetIbe an additive subgroup of 2. Then the right and left orders ofI are defined to be. O, (I) = {ge el Ig gl }, and O, (I) = {gE e[ gl gl }. We also define I" = {gE2iIgl gl}), the inverse ofL IfI is a right R-submodule of 2, then I is a (fractional) right R-ideal if I contains a regular element of 2, and if there is a regular. element d E 2 with dl g R. Left R-ideals are defmed sirnilarly. Following [AD], R is called right Pr(fer if for every finitely generated right Rideal I, IMiI= R, IIMi = O,(I). A left Prdfer ring is defined similarly. It is proved in [AD, (1.12)] that R is right Priifer if and only if it is left PrUfer. A ring is called right. (left) Bezout is any finitely generated right (Ieft) ideal is principal. We say that R is a. Dubrovin valuation ring if R is Bezout and RIJ(R) is a simple Artmian ring, where J(R) is the Jacobson radical ofR. A prime ideal P ofR is said to be localizable if C(P) = {c. E R 1 c is regular mod P} is an Ore set of R. Let P be a non-zero prime ideal of a PI. 36.

(40) PrUfer ring R. Then any element of C(P) is regular, C(P) is localizable and R. is a Dubrovin valuation ring ([D2]). We write Spec(R) for the set of all prime ideals ofR.. '. Lemma 5.1.1. Let S be an overring of R. Suppose that S isfl at as a left R-module. Then. S Q. S or Snaturally.. Proof1 For any oc =2s, X t,, where s,, t, E S, we define (p(oc) =Zs, t, . Then there is a regular element cE R with s, -- c-i s, for some s, E R. From the exact sequence O -->. s -> 2, we derive the exact sequence O -SQ. S- 2 Q. S• Then ct = Zs, Q t, = Zc-' s, X t,= c-iQ2 s, t, E 2 X. S. So if (p(oc) = O, then O= ]Z)s, t,= c-i(2 s, t,) and thus ct = O, which shows that rp is one-to-one. It is clear that rp is onto and hence q. is an isomorphism.. Let I be a right ideal ofR and s E2. Then we use the following notation; s'iI. ={r ERIsrEI} which isaright ideal ofR. ' Lemma 5.1.2. Under the same notation and assumption as in Lemma 5.1.1, letI be a non-zero right ideal of R and lets E S, non-zero. Then (s-iI)S = s-' (IS) = {t ESlst E IS}.. Pro<l: It is clear that (s-'I)Sg s-' (I S). To prove the converse inclusion, we consider. the exact sequence O -> s-il -> R4' S/I, where si (r) = [sr + I] for alt r ER. Then since S is a flat left R-module, we have the following exact sequence: O -> (s-il)Q. S -->R Q. S ---l-L' i!iXi ,S><ll X. S.. From the exact sequence we derive the following exact sequence: o- (s-iI)s --> s ----E-L' -> ,Sl!lll' S,. because (S/I) X. Sg (S X. S)/(I Q, S) cr Sl(IS) by Lemma 5.1.1, which shows that (s'iI)S2 s"i (I S). Hence (s-iI )S = s-i (I S) fo11ows.. 37.

(41) A family gr of right ideals of R is called a right Gabriel topology on R if gn satisfies the following two conditions: (i) ifl E gn and r E R, then r-il E ga, and. '. (ii) if IEgy and .I is aright ideal ofR such that a-iul E gn for alla EI , thenJd gn.. If gr is a right Gabriel topology on R, then we write R, for the right quotient of. R with respect to 7. SinceR is a prime Goldie ring, R, = u{(R:Di1I E ge}, where (R:Di. " {g E 2 1 gl g R}. We refer the readers to [S] for elementary properties of Gabriel topology.. Proposition 5.1.3. Let S be an overring of R. Suppose that S isflat as a left R-module. Then gr(E}) = {I: right ideal of R 1 IS = S } is a right Gabriel topology on R andS = Rgr(s).. Proof Let ,J E gn(S) and r E R (r 4 O). Then Rl(r'iD cr (rR +DII implies (r-ibS= S, i.e., r-'I E gr(S), because IS = S. Next, let I E gn(S) and let Jbe a right ideal ofR such. that (a-'.l)S =Sfor all a E L ThenS2JS 2 2.., a(a-i-1)S = 2].., aS= S• ThUS JS = S, i.e., J E gn(S). Hence gn(S) is a right Gabriel topology on R.. To show that S=' R,(,), let I E gn(,st. Then S = RS 2 (R:I)i IS ? (.R:Db which. ' inclusion, let s e S. Then S= s-iS= (s-iR)s implies Ry(,) g S To show the converse by Lemma 5.1.2 and so s-'R E gn(S) and s E (R: s-'R)ig&(s) Hence S= Rgr(s).. . Corollary 5.1.4. Under the same notation and assumptions as in Proposition 5.1.3, let I' be a right ideal of S. Then I' = (Z'n R) S.. Since any overring of a PrUfer ring R is flat as a right R-module as well as a left. R-module, we have. Corollary 5.1.5. Let S be an overring of a Profer ring R. Then there is a right aefP Gabriel topology gn (gy ' ) on R such that S == Rgr = Rgr) .. 38.

(42) 5.2. Prime ideals of overrings of a PI Prtifer ring. In this section, we assume that R is a PI PrUfer ring. Note that R. is a Dubrovin valuation ring for any P E Spec(R) and that any overring ofa PrUfer ring is PrUfer (see. [MMU, (2.6)]).. '. Lemma 5.2.1. Let P E Spec(R) and P, ' e Spec(R. ). Then P,= P, 'A R E Spec(R) and Rp, =(Rp)p,"'. Proof: Since J((R.).,,)AR. =P, ', we have P, =J((R. ).,,) rAR and so P, E Spec(R) by [Mo, (1.8)]. Since J( R. ) 2P, ', we have P= J( R. ) r> R2P, and so C(P) p C( Pi ). by [MMU, (17.1)]. To prove that C(P,) g C.,(P, ') ={cuER. Ia is regular mod P,'}, let cE C(P,)and c6 EP,'for some BE R,.Then there is adE C(P) and Bd E R, i.e., cPd E P, and so Bd E P,. Thus B E P,' and hence C(P,)g C,, (P,'). This implies that R., g (R. ).,, . To prove the converse inclusion, we claim that acE C( P, ). for any cuE C..(P, ') andcE C(P) with cxcE R. Assume that (xcr EP, for somerE. ' R. Then cr EP,' and so rG nyA R= P,,because C(P) g C( P,)gC., (ny)• Hence (xcE C(P, ). Now, letxE (R.).,,.ThenxB E R. for some BEC.,(P, ') and so xSc e. ' R for some cEC(P) with Pc E R. Since Pc E C(Pl ), we havex ERp,•Hence Rp,= (R. ).,, follows.. Lemma 5.2.2. LetS be an overring of R and letP' E Spec(S)• Then P ==P' AR E Spec(R) and R.' = Sp••. Proof Since P ==P' A R = J( S.,) A SA R = J( S.,) A R, it follows from [Mo, (1.8)] that P E Spec(R). Let c E C(P) and assume that cs E P' with s E S. Then there is an I. E gr(S) with xl gR and so cxl g P. Thus xl gP and so x E P', which implies C(P) g. Cs (P'). Hence Rp g Sp,. Since Rp is a Dubrovin valuation ring, there is a P,'e. 39.

(43) Spec(R. ) with S.,= (R. ).,, by Theorem 1.2.6. We put P, =J((R. ),,, )AR. Then P, = J( S.,) AR =P. Hence R. = R,, = (R. ).,, =S.,by Lemma 5.2.1.. Theorem 5.2.3. LetR be a PIProfer ring and letS be an overring of R. Then Spec(S) = {PS l P E Spec(R) with PS c S}and S = r'> R. , where P runs over all PE Spec(R). withPScS. ProqC Let P E Spec(R) with PS c S and let I' be a maximal right ideal of S with I' R PS. Then M' = ann,(S/I') = {s E S1 (S(I')s =O} is a maximal ideal ofS and so P, = M'. A R E Spec(R) with R., = S., by Lemma 5.2.2. Because of ,SJt(M' crS.,/(M'S., ) i 2(S/M'), the quotient ring of 51M', we havePS.,g I.,c S.,. So if P, P P, then R.,. =PR., =PS., cS.,, a contradiction Thus P, ?P and so R. p R,, 2Sby [MMU, (17.1)]. Hence S g AR. , where P runs over all PE Spec(R) with PS c S. Furthermore,. for any PE Spec(R) with PS c S, let P' = J(R. ) A S. Then P' E Spec(S) by [Mo, (1.8)]. Since P = J(R. ) AR =P' A S, it follows frorn Corollary 5.1.4 that P' = PS.. Hence PS G Spec(S). Conversely, let P' E Spec(S) with S )P'. Then P =P' A R E. Spec(R), P' = PS and R. = S.,by Corollary 5.1.4 and Lemna 5.2.2. Hence S = A R. by [MMU, (14.6)], where P runs over all PE Spec(R) with PS c S and Spec(S) =. {pE Spec(R) 1 pS c S}.. Corollary 5.2.4. LetR be a PI Pr4Cer ring and let P, E Spec(R) (i -- 1, 2)• Then Pi +. P2=Ror Pi ? P2 or Pi g P2•. Proof Suppose that R ? P,+P, . Let Mbe a maximal ideal of R with Mp P,+P,.. Then P, R. E Spec( R. ) by Theorem 5.2.3 and so either P, R. 2 P, R. or Pi R. gP, R. by Proposition 1.2.5. Sii}ce C(M) g C( P,) by [MMU, (17.1)], we have. P, RM AR= P,.Hence either Pi 2 P2 or Pi g P2•. 40.

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